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c) This compound has a DU of 4. The strong and broad absorption in the 3500-3200 cm⁻¹ region indicates that the compound is an alcohol. The absorptions in the regions of 3100-3000 cm⁻¹ are due to sp²-hybridized C-H bonds, and those at 3000-2850 cm⁻¹ are due to sp³-hybridized C-H bonds. The appearance of four bands in the 1600-1450 cm⁻¹ region and the strong bands in the 900-675 region suggest the presence of an aromatic ring. The NMR spectrum shows the presence of four different types of hydrogens and the integrals match the actual number of hydrogens. The signal at 7.3 δ, due to five CH₃ hydrogens, indicates the compound has a monosubstituted aromatic ring. The singlet at 2.7 δ is probably due to the HO-CH hydroxy hydrogen, which is not coupled due to rapid chemical exchange. The H that appears as a quartet at 4.8 δ must be coupled to the 3 H's appearing as a doublet at 1.4 δ (a CHCH₃ group). The unknown is 1-phenyl-1-ethanol. 14.13 The IR spectrum of this compound shows an OH band in the 3500-3200 cm⁻¹ region and a strong absorption in the C-O region. The absorptions in the region of 3000-2850 cm⁻¹ are due to sp³-hybridized C-H bonds. The IR data along with its DU of 0 indicate that this is a saturated alcohol. There are only four signals (all CH₂) in the spectrum indicating that the compound has symmetry. Three of the CH₂ groups must be duplicated. The CH₂ group at 63 δ must be attached to an oxygen. The absence of any CH₃ or CH groups indicates that the OH groups must be on the ends of the chain of carbons. 14.14 a) The DU is 0. The IR spectrum shows the presence of an OH group, and H's on sp³-hybridized C's, so the compound is a saturated alcohol. The coupling in the is complex and difficult to analyze. The singlet at 2.6 δ is probably due to the hydrogen of the OH, which is not coupled due to rapid chemical exchange. The spectrum is more useful in this case. It shows the presence of five different carbons. The CH₂ at 68 δ must be bonded to the because it is furthest downfield. Because its chemical shift is further downfield, the CH is probably bonded to this CH₂. This allows the CH₃ fragment CHCH₂OH to be written. Attaching the remaining fragments, a CH₂ and two different CH₃ groups, results in 2-methyl-1-butanol. 217